Find the path integral of $E(x)=-kq(\frac{x}{(x^2+y^2)^\frac{3}{2}},\frac{y}{(x^2+y^2)^\frac{3}{2}})$ along the straight line connecting $(1,0)$ and $(0,1)$ by calculating its line integral.
I really need to sanity check on this, I was able to find the integral to equal 0 by calculating it through the path of a quarter of the unit circle, since the electric field is path-independent, I should be getting 0 for this integral too. But I couldn't seem to get it right.
$\newcommand{\vect}[1]{{\bf #1}}$
Note that if you define $V(x,y)$ as
$$ V(x,y) = \frac{qk}{(x^2+y^2)^{1/2}} $$
Then
$$ \vect{E}(x,y) = -\nabla V(x,y) $$
So in general the integral along a path connecting the points $\vect{a}$ and $\vect{b}$ is
$$ \int_{\vect{a}}^{\vect{b}}{\rm d}\vect{l}\cdot \vect{E} = -\int_{\vect{a}}^{\vect{b}}{\rm d}\vect{l}\cdot\nabla V = -(V(\vect{b}) - V(\vect{a})) = V(\vect{a}) - V(\vect{b}) $$
So, you just need to evaluate the function $V$ at the end points of the path and calculate the difference.
NOTE: If the path is closed then $V(\vect{a}) = V(\vect{b})$ and the integral is zero